Last time… Resistors Circuits Today…€¦ · Tue. Oct. 13, 2009 Physics 208 Lecture 12 1 Last...

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Tue. Oct. 13, 2009 Physics 208 Lecture 12 1

Last time…   Begin circuits

  Resistor circuits   Start resistor-capacitor circuits

Today…


Resistors

Schematic layout

Circuits

Physical layout


Quick Quiz Which bulb is brighter?

A. A

B. B

C. Both the same

Current through each must be same

Conservation of current (Kirchoff’s current law)

Charge that goes in must come out

I

I

I

I

Question When current flows, charge moves around the circuit.

Suppose that positive charge carriers flow around the circuit. What is the change in potential energy of a positive charge as moves from c to d?


A. qVd – qVc

B.  qVc – qVd

C. qVd + qVc

D.  zero

Quick Quiz

What is the change in kinetic energy as it moves from c to d?


A. qVd – qVc

B.  qVc – qVd

C. qVd + qVc

D.  zero


Power dissipation (Joule heating)   Charge loses energy from c to d.

  Ohm’s law:  

 

  Energy dissipated in resistor as   Heat (& light) in bulb

  Power dissipated in resistor =

€

Elost = −ΔE = − ΔKE + ΔU( ) = 0 − q Vd −Vc( )

€

Vc −Vd( ) = IR

€

dElost

dt=dqdtIR = I2R Joules / s = Watts

€

Elost = qIR

2


Light bulbs and power   Household voltage is 120V

     

  Cost   24 hours on requires

  MG&E ~ 13¢ / kWatt-hour

60 Watt

€

60W = 60J /s = I2R = I IR( ) =VI

€

I = 60W /120V = 0.5A

€

60J /s( ) 24hour( ) 3600s /hour( ) = 5,184,000J

€

1kW − hour = 1000J /s( ) 3600s /hour( ) = 3,600,000J

€

R =V /I =120V /0.5A = 240Ω

19¢ / day Tue. Oct. 13, 2009 Physics 208 Lecture 12 8

Two different bulbs

  Current same through each   Power dissipated different

  Brightness different

R1

R2

a

b

c

d

e

I

I

I

I €

P1 = I2R1

€

P2 = I2R2


Kirchoff’s junction law

  Charge conservation

Iin

Iout

Iout = Iin

I1

I2

I3 I1=I2+I3

I2

I3

I1

I1+I2=I3


Quick Quiz What happens to the brightness of the

bulb A when the switch is closed?

A.  Gets dimmer

B.  Gets brighter

C.  Stays same

D.  Something else


Question As more and more resistors are added to the

parallel resistor circuit shown here the total current flowing I…

…. R1 R2 R3 R4

I

A.  Increases if each Ri getting bigger B.  Increases if each Ri getting smaller C.  Always increases D.  Always decreases E.  Stays the same

Each resistor added adds ΔV/Ri to the total current I


You use one power strip to plug in your toaster, coffee pot, microwave.

Toaster Coffee Pot Microwave

10 A 5 A 12 A

Everything works great until you plug in your space heater, then you smell smoke. This is because

Question

A.  The resistance of the circuit is too high

B.  The voltage in the circuit is too high

C.  The current in the circuit is too high

3


More complicated circuits

  Both series & parallel   Determine equivalent

resistance   Replace combinations

with equivalent resistance


Quick Quiz The circuit below contains three 100W light bulbs. The emf ε = 110 V. Which light bulb(s) is(are) brightest ?

A. A B. B C. C D. B and C E. All three are equally bright.


Measurements in a circuit   A multimeter can measure currents (as an ammeter), potential

difference (as a voltmeter)   Electrical measuring devices must have minimal impact in the

circuit

R

ε

V Voltmeter

The internal resistance of the ammeter must be very small I = IA ε = ΔV+ΔVA = RI + rAI → RI for rA →0

ΔVA A

R

ε Ammeter

I IA

ΔV

The internal resistance of the voltmeter must be very large I = Iv+IR ΔVV = ε

ΔVV

IV

IR

I

€

I =εrV

+εR

rV →∞ → εR Tue. Oct. 13, 2009 Physics 208 Lecture 12 16

Kirchoff’s loop law

  Conservation of energy

R1

R2 R3 ε

I1

I2 I3

Thur. Oct. 16, 2008 Physics 208 Lecture 14 17

Resistor-capacitor circuit   What happens to charges on

the capacitor after switch is closed?

  Why does the charge on the capacitor change in time?

  Why does the charge flow through the resistor?


Charging a capacitor   Again

Kirchoff’s loop law:

€

ε − IR −QC /C = 0

Time t = 0:

€

Qc = 0⇒ I =ε /R€

⇒ I =ε /R −QC /RC

Looks like resistor & battery: uncharged cap acts like short circuit

t increases:

€

Qc > 0⇒ I <ε /R VC increases, so VR decreases

Time t = ∞:

€

VC =ε ⇒VR = 0⇒ I = 0 Fully charged capacitor acts like open circuit

4


Discharging the capacitor

  Kirchoff’s loop law

A

B C

D

€

VB −VA( ) + VD −VC( ) = 0

€

ΔVc =Qc /C

€

−IR

€

⇒ I =Qc

RCCharges in the current I come from capacitor:

€

I = −dQc

dtThur. Oct. 16, 2008 Physics 208 Lecture 14 20

RC discharge

  RC time constant

€

τ = RC

€

Q =Qoe−t /τ

€

I = Ioe−t /τ


Charging a capacitor

€

Q =Qmax 1− e−t /τ( )


Question The circuit contains three identical light

bulbs and a fully-charged capacitor. Which is brightest?

A.  A

B.  B

C.  C

D.  A & B

E.  All equally bright


Question The circuit contains three identical light

bulbs and an uncharged capacitor. Which is brightest?

A.  A

B.  B

C.  C

D.  A & B

E.  All equally bright


RC discharge   RC time constant

time t

€

τ = RC

5


RC analysis

  Kirchoff loop law:

€

ΔVC + ΔVR = 0

⇒QC

C− IR = 0

  I related to QC

€

I = −dQC

dt

€

QC

C+ R dQC

dt= 0

€

dQC

dt= −

QC

RCTue. Oct. 13, 2009 Physics 208 Lecture 12 26

RC analysis

€

dQC

dt= −

QC

RC

€

dQC

QC

= −1RC

dt

dQC

QC

= −1RC

dt0

t

∫Qo

Q

∫

lnQC Qo

Q t( ) = −tRC

QC t( ) =Qo exp −t /RC( )

Last time… Resistors Circuits Today…€¦ · Tue. Oct. 13, 2009 Physics 208 Lecture 12 1 Last...

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Transcript of Last time… Resistors Circuits Today…€¦ · Tue. Oct. 13, 2009 Physics 208 Lecture 12 1 Last...